3.492 \(\int \frac{\tanh (e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=69 \[ \frac{1}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]
[Out]
-(ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(3/2)*f)) + 1/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^
2])
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Rubi [A] time = 0.0738817, antiderivative size = 69, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used =
{3194, 51, 63, 208} \[ \frac{1}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{f (a-b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]
[Out]
-(ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a - b]]/((a - b)^(3/2)*f)) + 1/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^
2])
Rule 3194
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\tanh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) (a+b x)^{3/2}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=\frac{1}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 (a-b) f}\\ &=\frac{1}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^2(e+f x)}\right )}{(a-b) b f}\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a-b}}\right )}{(a-b)^{3/2} f}+\frac{1}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}
Mathematica [C] time = 0.0781408, size = 58, normalized size = 0.84 \[ -\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \cosh ^2(e+f x)}{a-b}+1\right )}{f (b-a) \sqrt{a+b \cosh ^2(e+f x)-b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tanh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]
[Out]
-(Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Cosh[e + f*x]^2)/(a - b)]/((-a + b)*f*Sqrt[a - b + b*Cosh[e + f*x]^2]
))
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Maple [C] time = 0.123, size = 93, normalized size = 1.4 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ( -{\frac{\sinh \left ( fx+e \right ) }{-{b}^{2} \left ( \sinh \left ( fx+e \right ) \right ) ^{6}+ \left ( -2\,ab-{b}^{2} \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{4}+ \left ( -{a}^{2}-2\,ab \right ) \left ( \sinh \left ( fx+e \right ) \right ) ^{2}-{a}^{2}}\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x)
[Out]
`int/indef0`(-sinh(f*x+e)*(a+b*sinh(f*x+e)^2)^(1/2)/(-b^2*sinh(f*x+e)^6+(-2*a*b-b^2)*sinh(f*x+e)^4+(-a^2-2*a*b
)*sinh(f*x+e)^2-a^2),sinh(f*x+e))/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate(tanh(f*x + e)/(b*sinh(f*x + e)^2 + a)^(3/2), x)
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Fricas [B] time = 2.67186, size = 3475, normalized size = 50.36 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/2*((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^
2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f
*x + e) + b)*sqrt(a - b)*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x + e)^3 + b*sinh(f*x + e)^4 + 2*(4
*a - 3*b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - 3*b)*sinh(f*x + e)^2 + 4*sqrt(2)*sqrt(a - b)*sqrt((
b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x +
e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - 3*b)*cosh(f*x + e))*sinh(f*x + e) + b)
/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x + e)^2 + 1)*sinh(f*x + e
)^2 + 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 + cosh(f*x + e))*sinh(f*x + e) + 1)) - 4*sqrt(2)*((a - b)*cosh(f*
x + e) + (a - b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*co
sh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2)))/((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^4 + 4*(a^2*b - 2*a*b^2
+ b^3)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a^2*b - 2*a*b^2 + b^3)*f*sinh(f*x + e)^4 + 2*(2*a^3 - 5*a^2*b + 4*a
*b^2 - b^3)*f*cosh(f*x + e)^2 + 2*(3*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^2 + (2*a^3 - 5*a^2*b + 4*a*b^2 -
b^3)*f)*sinh(f*x + e)^2 + (a^2*b - 2*a*b^2 + b^3)*f + 4*((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^3 + (2*a^3 -
5*a^2*b + 4*a*b^2 - b^3)*f*cosh(f*x + e))*sinh(f*x + e)), -((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x +
e)^3 + b*sinh(f*x + e)^4 + 2*(2*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^2 + 4
*(b*cosh(f*x + e)^3 + (2*a - b)*cosh(f*x + e))*sinh(f*x + e) + b)*sqrt(-a + b)*arctan(-1/2*sqrt(2)*sqrt(-a + b
)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + si
nh(f*x + e)^2))/((a - b)*cosh(f*x + e) + (a - b)*sinh(f*x + e))) - 2*sqrt(2)*((a - b)*cosh(f*x + e) + (a - b)*
sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(
f*x + e) + sinh(f*x + e)^2)))/((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^4 + 4*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*
x + e)*sinh(f*x + e)^3 + (a^2*b - 2*a*b^2 + b^3)*f*sinh(f*x + e)^4 + 2*(2*a^3 - 5*a^2*b + 4*a*b^2 - b^3)*f*cos
h(f*x + e)^2 + 2*(3*(a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^2 + (2*a^3 - 5*a^2*b + 4*a*b^2 - b^3)*f)*sinh(f*x
+ e)^2 + (a^2*b - 2*a*b^2 + b^3)*f + 4*((a^2*b - 2*a*b^2 + b^3)*f*cosh(f*x + e)^3 + (2*a^3 - 5*a^2*b + 4*a*b^2
- b^3)*f*cosh(f*x + e))*sinh(f*x + e))]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh{\left (e + f x \right )}}{\left (a + b \sinh ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)**2)**(3/2),x)
[Out]
Integral(tanh(e + f*x)/(a + b*sinh(e + f*x)**2)**(3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tanh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate(tanh(f*x + e)/(b*sinh(f*x + e)^2 + a)^(3/2), x)